consider a head on collision between two particles of masses `m_1 and m_2`. The initial speeds of the particles are `u_1 and u_2` in the same direction. The collision starts at t=0 and the particles interact for a time interval `/_\t.` during the collision the speed of the first particle varies as
`v(t)=u_1+t//_\ t(v_1-u_1)` Find the speed of the second particle as as function of time during the collision.

Explanation

Using lw of conservation of momentum `m_1u_1+m_2u_2=m_1v(t)+m_2v^1`
where v=speed of 2nd particle during collision
`rarr m_1u_1+m_2u_2`
1=m_1u_1+m_1.(t//_\)(v_1-u_1)+m_2v^1`
`rarr (m_2u_20/m_2=m_1/m_2(t//_\t)(v_1-u_1)+v^1`
`: v^1=u_1-m_1/m_2 t//_\t(v_1-u_1)`